MATHS 3 Marks Question

Plotting the points P (1,1), 0 (2, -3) and R (-1, -2) on the graph paper and join these three points, we get three lines i.e., the given points do not lie on the same line. So, given points are not collinear.

Thinking Process:

• Firstly, draw the perpendicular lines from the point to the coordinates axes.
• Further, measure the distance from intersecting points to the origin along their sign.
• Finally, write the x unit distance and y unit distance in pair.

Sol. The coordinates of the points P, Q, R, S, T and O are P(1, 1), Q(-3, 0), R(-2, -3), S(2, 1), T (4, -2) and O(0, 0).

Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3, 2) lies in II quadrant, Q(-7, -3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2, 2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

In point P(1, 0), y-coordinate is zero, so it lies on X-axis. In point Q(4, 0), y-coordinate is zero so it lies on X-axis. In point S(1, 3), both coordinates are positive, so it lies in I quadrant. On plotting these points, we get the following graph.

Now, take a point R on the graph such that PQRS is a square. Then, all sides will be equal i.e., PQ = QR = RS = PS. So, abscissa of R should be equal to abscissa of Q i.e., 4 and ordinate of R should be equal to ordinate of S i.e., 3. Hence, the coordinates of R are (4, 3).

Here, in point 4(1, 3) both x and y-coordinates are positive, so it lies in I quadrant. In point 8(-3, -1), both x and y coordinates are negative, so it lies in III quadrant. In point C(1, -4), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant.
In point D(-2, 3), x-coordinate is negative and y-coordinate is positive, so it lies in II quadrant. In point E(0, -8) x-coordinate is zero, so it lies on Y-axis and in point F(1, 0), y-coordinate is zero, so it lies on X-axis.
On plotting the given points, we get the following graph.

Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, -0.25) on the graph paper.

Given length of rectangle = 5 units
Given breadth of rectangle = 3 units
One vertex is at origin i.e.., D(0, 0) and one of the other vertices lies in III quadrant. So the Length of the rectangle is 5 units in the negative direction of X–axis and then vertex is A(-5, 0).Also the breadth of the rectangle is 3 units in the negative direction of Y-axis and then vertex is C(0, -3). The fourth vertex B is (5, -3).

Plotting the points 0 (0, 0), A (2, 2)and 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative direction of X-axis, then its x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

1. The coordinates of the point which lies on both the axes are (0, 0).
2. The coordinates of the point whose ordinate is -4 and which lies on y-axis are (0, -4).
3. The coordinate of the point whose abscissa is 5 and which lies on x-axis are (5, 0).

1. Clearly, the distance of points A, L and O from y-axis is 0. So, A(0, 3), L(0, -4) and O(0, 0) are the points whose abscissa is 0.
2. Clearly, the distance of points G, I and O from x-axis is 0. So, G(5, 0), I(-2, 0) and O(0, 0) are the points whose ordinate is 0.
3. Clearly, the distance of points H and D from y-axis is 5 units and both lien in second and third quadrants respectively. So, (-5, -3) and D(-5, 1) are the points whose abscissa is -5.

Plotting the points P (1, 3), Q (-1, -1) and R (-2, -3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear.